Commercial Mathematics

Compound interest

Compound interest models exponential growth of capital where interest is reinvested periodically, governed by discrete exponential sequences or continuous differential equations. This topic covers its algebraic formulation and logarithmic manipulation.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

📖 Topic progress

○ Mark as complete

Definition and statement

Compound Interest Growth Principle

If a principal $P$ is invested at an annual nominal rate of interest $r$ compounded $n$ times per year for $t$ years, the accumulated amount $A$ is given by the discrete exponential function $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$

📐 Geometric Interpretation

Geometrically, compound interest represents an exponential curve $y = P(1 + r/n)^{nt}$ that exhibits monotonic non-linear growth with a strictly positive and increasing slope, contrasting with the linear growth characteristic of simple interest.

⚙️ Algebraic & Structural Meaning

Algebraically, the relationship forms a geometric progression where the growth factor per conversion period is $1 + \frac{r}{n}$. Taking logarithms reduces multiplicative compounding to linear equations, facilitating quick analytical solutions for time or rate unknowns.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Constant Nominal Interest Rate	`Entire investment duration $[0, t]$`	Strictly Mandatory	If the rate varies over time, the standard formula $A = P(1 + r/n)^{nt}$ fails, and one must integrate or multiply discrete variable growth factors: $A = P \prod_{i=1}^{nt} (1 + \frac{r_i}{n})$.y
Regular Conversion Periods	`Equally spaced discrete intervals $\Delta t = \frac{1}{n}$`	Strictly Mandatory for standard formula	Irregular compounding intervals invalidate the uniform exponent $nt$, requiring fractional or custom compounding formulas for each specific sub-interval.
Non-negative Principal and Rates	$P > 0, r \ge 0$	Essential for financial validity	Negative principal or negative absolute interest rates violate standard commercial definitions, leading to negative amounts or absurd exponential decay where none is intended.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Doubling Time Rule of 72 / 69.3

$$t \approx \frac{72}{\text{Rate}(\%)} \quad \text{or} \quad t \approx \frac{0.693}{\ln(1+r)}$$

Rule: To find approximate time for money to double at compound interest, divide 72 (or 69.3 for continuous compounding) by the percentage interest rate.

Application: Use this to bypass tedious logarithmic calculations when estimating doubling periods.

Multi-Year Amount Ratio Property

$$\frac{A_{t+1}}{A_t} = 1 + r, \quad \frac{A_{t+k}}{A_t} = (1 + r)^k$$

Rule: Amounts accumulated at successive equal time intervals form a geometric progression. The ratio of amounts after $(t+k)$ years to $t$ years depends solely on $k$, independent of $P$.

Application: Standard pattern in KVS and DSSSB where a sum becomes $x$ times in $a$ years and $y$ times in $b$ years.

Difference Between CI and SI for 2 Years

$$CI - SI = P \left(\frac{r}{100}\right)^2$$

Rule: Direct formula to find principal or rate when the 2-year compound interest and simple interest difference is given.

Difference Between CI and SI for 3 Years

$$CI - SI = P \left(\frac{r}{100}\right)^2 \left(3 + \frac{r}{100}\right)$$

Rule: Immediate extraction of principal or rate for 3-year differential problems without expanding full binomials.

Fractional Time Compounding

$$A = P \left(1 + \frac{r}{100}\right)^a \left(1 + \frac{b}{c} \frac{r}{100}\right)$$

Rule: For time $t = a \frac{b}{c}$ years (where $a$ is integer and $b/c$ is fraction), compound for $a$ years at full rate and simple interest for the fractional remainder.

Rapid Elimination via Unit Digits / Modulo Arithmetic

$$A \equiv P(1 + r)^t \pmod{10}$$

Rule: Examine the last digit of the calculated amount using modular arithmetic to eliminate 2 incorrect options instantly.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Nominal vs. Effective Annual Rate Confusion

Common Pitfall: Candidates use the nominal rate $r$ directly in annual formulas when compounding occurs semi-annually, quarterly, or monthly without adjusting rate to $r/n$ and time to $nt$.

Counterexample: $r = 12\%$ compounded quarterly for $1$ year calculated as $P(1+0.12)^1$ instead of $P(1+0.03)^4$.

Why it fails: Failing to adjust conversion frequency ignores the higher effective yield generated by more frequent compounding.

🛡️ Exam Defense: Always scale the rate by dividing by conversion frequency $n$ and multiply the time exponent by $n$ before calculation.

⚠️ COMMON ERROR #02

Simple vs. Compound Interest Formula Overlap

Common Pitfall: Applying linear simple interest formulas ($A = P(1 + rt)$) to compound interest problems or vice versa.

Counterexample: Calculating 3-year compound growth by multiplying rate by 3 ($3r$) instead of compounding factor $(1+r)^3$.

Why it fails: Compound interest incorporates interest-on-interest, making linear approximations underestimate true growth significantly for $t > 1$.

🛡️ Exam Defense: Check the keyword carefully: 'compounded annually/half-yearly' strictly demands exponential factors.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Simple Interest (SI)	Interest calculated on initial principal only	$I = \frac{P \cdot r \cdot t}{100}$, $A = P(1 + rt)$	Straight line with constant slope $Pr/100$	Keywords: 'Simple interest', constant annual increment
Compound Interest (Discrete)	Interest added to principal at fixed intervals	$A = P \left(1 + \frac{r}{100n}\right)^{nt}$	Exponential curve with increasing slope	Keywords: 'Compounded annually / half-yearly / quarterly'
Continuous Compounding	Compounding frequency approaches infinity ($n \to \infty$)	$A = P e^{rt}$	Smooth exponential growth curve governed by Euler's constant	Keywords: 'Compounded continuously', instantaneous growth
Depreciation (Retardation)	Exponential decay over successive time periods	$A = P \left(1 - \frac{r}{100}\right)^t$	Downward sloping exponential decay curve	Keywords: 'Depreciates', 'decreases at annual rate of'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

A sum of money placed at compound interest triples itself in 4 years. In how many years will it become 27 times itself?

(A) 8 years(B) 12 years(C) 16 years(D) 24 years

📚 Method A: Formal Textbook Derivation

Let principal be $P$. Given $3P = P(1+r)^4 \implies (1+r)^4 = 3$. We need amount to be $27P$: $27P = P(1+r)^t \implies 3^3 = (1+r)^t$. Substituting $(1+r)^4 = 3$: $((1+r)^4)^3 = (1+r)^t \implies (1+r)^{12} = (1+r)^t \implies t = 12$.

⚡ Method B: 25-Second Exam Shortcut

Since $27 = 3^3$, multiply the base time of 4 years by 3 to get $4 \times 3 = 12$ years instantly.

💡 Tip: Look for powers of the multiplier: if it becomes $x^k$ times, the time is $k \times$ original time.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The difference between compound interest and simple interest on a certain sum of money at $5\%$ per annum for 2 years is Rs. 25. Find the principal.

(A) Rs. 5,000(B) Rs. 10,000(C) Rs. 20,000(D) Rs. 25,000

📚 Method A: Formal Textbook Derivation

Using the 2-year difference formula: $CI - SI = P \left(\frac{r}{100}\right)^2$. Given $25 = P \left(\frac{5}{100}\right)^2 = P \left(\frac{1}{20}\right)^2 = \frac{P}{400}$. Thus $P = 25 \times 400 = 10,000$.

⚡ Method B: 25-Second Exam Shortcut

Apply shortcut $P = \frac{\text{Diff} \times 100^2}{r^2} = \frac{25 \times 10000}{25} = 10,000$.

💡 Tip: Memorize the 2-year and 3-year difference formulas to solve.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If a sum amounts to $ \text{Rs. } 4500 $ in 2 years and $ \text{Rs. } 6750 $ in 4 years at compound interest, what is the principal?

A Rs. 2,000

B Rs. 3,000

C Rs. 3,500

D Rs. 4,000

Derivation & Analysis:

Using geometric property $ \frac{A_4}{A_2} = \frac{6750}{4500} = 1.5 = 1+r $. Principal $ P = \frac{A_2}{(1+r)^2} = \frac{4500}{(1.5)^2} = \frac{4500}{2.25} = \text{Rs. } 3,000 $.

MCQ 2: The compound interest on $ \text{Rs. } 10,000 $ at $ 10\% $ per annum compounded half-yearly for 1 year is:

A Rs. 1,000

B Rs. 1,025

C Rs. 1,050

D Rs. 1,100

Derivation & Analysis:

Half-yearly rate $ r' = 5\% $, time periods $ n = 2 $. Amount $ A = 10000 (1 + \frac{5}{100})^2 = 10000 (1.05)^2 = 10000 \times 1.1025 = 11025 $. Interest $ CI = 11025 - 10000 = \text{Rs. } 1025 $.

MCQ 3: At what annual rate of compound interest will $ \text{Rs. } 1600 $ amount to $ \text{Rs. } 2500 $ in 2 years?

A 15\%

B 20\%

C 25\%

D 30\%

Derivation & Analysis:

Formula: $ 2500 = 1600 (1 + \frac{r}{100})^2 \implies \frac{25}{16} = (1 + \frac{r}{100})^2 \implies \frac{5}{4} = 1 + \frac{r}{100} \implies \frac{r}{100} = \frac{1}{4} \implies r = 25\% $.

MCQ 4: A machine depreciates in value at the rate of $ 10\% $ per annum. If its present value is \text{Rs. } 81,000, what was its value 2 years ago?

A Rs. 90,000

B Rs. 95,000

C Rs. 1,00,000

D Rs. 1,05,000

Derivation & Analysis:

Value 2 years ago $ P $ satisfies $ 81000 = P (1 - \frac{10}{100})^2 = P (0.9)^2 = 0.81 P \implies P = \frac{81000}{0.81} = \text{Rs. } 1,00,000 $.

MCQ 5: The difference between simple and compound interest on a sum for 3 years at $ 10\% $ per annum is \text{Rs. } 31. The sum is:

A Rs. 500

B Rs. 800

C Rs. 1,000

D Rs. 1,200

Derivation & Analysis:

Formula for 3 years: $ CI - SI = P \left(\frac{r}{100}\right)^2 (3 + \frac{r}{100}) \implies 31 = P (\frac{1}{10})^2 (3 + \frac{1}{10}) = P (\frac{1}{100}) (\frac{31}{10}) = \frac{31 P}{1000} \implies P = \text{Rs. } 1,000 $.

MCQ 6: In how many years will $ \text{Rs. } 8,000 $ amount to $ \text{Rs. } 9,261 $ at $ 5\% $ per annum compound interest?

A 2 years

B 3 years

C 4 years

D 5 years

Derivation & Analysis:

$ 9261 = 8000 (1 + \frac{5}{100})^t \implies \frac{9261}{8000} = (\frac{21}{20})^t \implies (\frac{21}{20})^3 = (\frac{21}{20})^t \implies t = 3 $ years.

MCQ 7: If the effective rate of interest is $ 10.25\% $ per annum corresponding to a nominal rate compounded semi-annually, the annual nominal rate is:

A 5\%

B 10\%

C 20\%

D 10.25\%

Derivation & Analysis:

Effective rate $ r_{eff} = (1 + \frac{r}{2})^2 - 1 = 0.1025 \implies (1 + \frac{r}{2})^2 = 1.025^2 \implies 1 + \frac{r}{2} = 1.05 \implies \frac{r}{2} = 0.05 \implies r = 10\% $.

MCQ 8: Assertion (A): Compound interest grows faster than simple interest after the first conversion period. Reason (R): Compound interest is calculated on the accumulated amount of previous periods.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C Assert (A) is true but Reason (R) is false

D Assert (A) is false but Reason (R) is true

Derivation & Analysis:

By definition, compound interest reinvests prior interest, generating exponential growth which always outpaces linear simple interest growth for $ t > 1 $.

MCQ 9: A sum amounts to $ \text{Rs. } 2400 $ in 3 years and $ \text{Rs. } 3000 $ in 4 years at compound interest. The annual rate of interest is:

A 15\%

B 20\%

C 25\%

D 33.33\%

Derivation & Analysis:

Interest for the 4th year $ = 3000 - 2400 = 600 $. This interest accrued on the amount at year 3 $ (2400) $. Rate $ r = \frac{600}{2400} \times 100 = \frac{1}{4} \times 100 = 25\% $.

Q10

MCQ 10: Under continuous compounding, a principal $ P $ doubles in 10 years. What is the approximate continuous interest rate $ r $?

A 5\%

B 6.93\%

C 7.2\%

D 10\%

Derivation & Analysis:

Formula: $ 2P = P e^{10r} \implies \ln 2 = 10r \implies 0.6931 = 10r \implies r = 0.06931 \text{ or } 6.93\% $.

Common questions

Compound interest exhibits exponential growth because interest is added to the principal at each conversion period, earning further interest. Simple interest remains strictly linear since interest is computed solely on the fixed initial principal.

Failing to adjust the nominal rate and time period when compounding frequency is semi-annual, quarterly, or monthly (e.g., forgetting to divide rate by 2 and multiply time by 2 for semi-annual compounding).

Candidates should leverage the ratio property $ A_{t+k} / A_t = (1+r)^k $, the Rule of 72 for doubling periods, and modular arithmetic to eliminate incorrect options rapidly without full logarithmic expansion.